const uid = function(){
        return Date.now().toString(36) + Math.random().toString(36).substr(2);
    }

 <ul className='rating'>
    {Array.from({ length: 10 }, (_, i) => (
      <li key={`rating-${i + 1}`}>
        <input
          type='radio'
          id={`num${i + 1}`}
          name='rating'
          value={i + 1}
          onChange={handleChange}
          checked={selected === i + 1}
        />
        <label htmlFor={`num${i + 1}`}>{i + 1}</label>
      </li>
    ))}
  </ul>

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
        dummy = ListNode(0, head)
        left = dummy 
        right = head #we actually want right to be head + n 
        
        #setting right to be head + n 
        while n > 0 and right: 
            right = right.next 
            n -= 1 
        
        #this should get left at previous and right at after thanks to dummy node we inserted 
        while right: 
            left = left.next
            right = right.next 
        
        #delete node 
        left.next = left.next.next 
        
        return dummy.next 



#bottom solution is what I first came up with and works in leetcode except for if head = [1] and n = 1 which is the case I tried to account for in the second elif statement but it gave me an error saying cannot return [] even though it was fine for the first if statement 

#big O: is still O(n) just like 1st scenario 

class Solution: 
	def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
    	if head is None: 
            return []
        elif head.next is None and n == 1: 
            return []
        else: 
            current = head
            counter = 0 #will become length of LL 
            
            #to determine length of LL 
            while current is not None: 
                counter += 1 
                current = current.next 
                
            #consider adding an edge case here if n > counter 
            
            #second pass now knowing length of LL 
            previous, current = head, head 
            new_counter = 0 
            target = counter - n 
            while new_counter is not target + 1: 
                if new_counter == (target - 1): 
                    previous = current 
                if new_counter == target: 
                    after = current.next 
                    previous.next = after 
                    current.next = None 
                    break 


                current = current.next 


                new_counter += 1

            return head